Metamath Proof Explorer


Theorem topdifinffin

Description: Part of Exercise 3 of Munkres p. 83. The topology of all subsets x of A such that the complement of x in A is infinite, or x is the empty set, or x is all of A , is a topology only if A is finite. (Contributed by ML, 17-Jul-2020)

Ref Expression
Hypothesis topdifinf.t ⊢ T = x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A
Assertion topdifinffin ⊢ T ∈ TopOn ⁡ A → A ∈ Fin

Proof

Step Hyp Ref Expression
1 topdifinf.t ⊢ T = x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A
2 difeq2 ⊢ x = y → A ∖ x = A ∖ y
3 2 eleq1d ⊢ x = y → A ∖ x ∈ Fin ↔ A ∖ y ∈ Fin
4 3 notbid ⊢ x = y → ¬ A ∖ x ∈ Fin ↔ ¬ A ∖ y ∈ Fin
5 eqeq1 ⊢ x = y → x = ∅ ↔ y = ∅
6 eqeq1 ⊢ x = y → x = A ↔ y = A
7 5 6 orbi12d ⊢ x = y → x = ∅ ∨ x = A ↔ y = ∅ ∨ y = A
8 4 7 orbi12d ⊢ x = y → ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A ↔ ¬ A ∖ y ∈ Fin ∨ y = ∅ ∨ y = A
9 8 cbvrabv ⊢ x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A = y ∈ 𝒫 A | ¬ A ∖ y ∈ Fin ∨ y = ∅ ∨ y = A
10 1 9 eqtri ⊢ T = y ∈ 𝒫 A | ¬ A ∖ y ∈ Fin ∨ y = ∅ ∨ y = A
11 10 topdifinffinlem ⊢ T ∈ TopOn ⁡ A → A ∈ Fin